Question

Find the greatest number that will divide 132,242and 382 leaving a remainder 2 in each case.​

Answer 1

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• They leave reminder two, Hence we've to first subtract them from Two.

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Now,

• 132 – 2 = 130

• 242 – 2 = 240

• 382 – 2 = 380

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Now,

$\binom{ \frac{130 \: = {2}^{1} \times {3}^{0} \times {5}^{1} \times {13}^{1} \times {19}^{0} }{240 \: = {2}^{4} \times {3}^{1} \times {5}^{1} \times {13}^{0} \times {19}^{0} } ^{} }{ \frac{380 = {2}^{2} \times {3}^{0} \times {5}^{1} \times { {13}^{0} \times {19}^{1} }^{ } }{HCF \: = {2}^{1} \times {3}^{0} \times {5}^{1} \times {13}^{0} \times {19}^{0} } }$

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• Refer to the Attached image to know the way I have used to find the HCF.

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• Basically, HCF = Product of the smallest power of each common
• prime factor in the numbers.

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( Note — X⁰ = 1 )

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Now, HCF

• = 2¹ × 3⁰ × 5¹ × 13⁰ × 19⁰

• = 2 × 1 × 5 × 1 × 1

• = 10

Hence, HCF ( 130, 240, 380 ) = 10

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Therefore, 10 is the greatest number that will divide 132, 242 and 382 leaving a remainder 2 in each case.

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